Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{-5x + 25}{x + 1} \div \dfrac{x^2 - 7x + 10}{x + 1} $
Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-5x + 25}{x + 1} \times \dfrac{x + 1}{x^2 - 7x + 10} $ First factor the quadratic. $r = \dfrac{-5x + 25}{x + 1} \times \dfrac{x + 1}{(x - 5)(x - 2)} $ Then factor out any other terms. $r = \dfrac{-5(x - 5)}{x + 1} \times \dfrac{x + 1}{(x - 5)(x - 2)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -5(x - 5) \times (x + 1) } { (x + 1) \times (x - 5)(x - 2) } $ $r = \dfrac{ -5(x - 5)(x + 1)}{ (x + 1)(x - 5)(x - 2)} $ Notice that $(x + 1)$ and $(x - 5)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -5\cancel{(x - 5)}(x + 1)}{ (x + 1)\cancel{(x - 5)}(x - 2)} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $r = \dfrac{ -5\cancel{(x - 5)}\cancel{(x + 1)}}{ \cancel{(x + 1)}\cancel{(x - 5)}(x - 2)} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $r = \dfrac{-5}{x - 2} ; \space x \neq 5 ; \space x \neq -1 $